T:RC

RC circuit

A resistor–capacitor circuit(RC circuit), or RC filter or RC network, is an electric circuit composed of resistors and capacitors driven by a voltage or current source. The 1st order RC circuit composed of one resistor and one capacitor, is the simplest example of an RC circuit.

Natural response

The simplest RC circuit is a capacitor and a resistor in series. When a circuit composes of only a charged capacitor and a resistor, then the capacitor would discharge its energy into the resistor. This voltage across the capacitor over time could be found through KCL, where the current coming out of the capacitor must equal the current going through the resistor. This results in the linear differential equation

Cfrac{dV}{dt} + frac{V}{R}=0
.

When solved, it results in the exponential decay function

V(t)=V_0 e^{-frac{t}{RC}}

Complex impedance

The equivalent resistance of a capacitor increases in relation to the amount of charge stored on the capacitor. If a capacitor is subjected to an alternating current voltage source, then the voltage of the capacitor would flip to the frequency of the AC voltage source. The faster the voltage of the AC voltage source flips, the less time charge would allowed to be stored on the capacitor, therefore reducing the capacitor's equivalent resistance. This explains the inverse relationship the equivalent resistance of a capacitor has with the frequency of the voltage source.

Frequency-domain considerations

These are frequency domain expressions. Analysis of them will show which frequencies the circuits (or filters) pass and reject. This analysis rests on a consideration of what happens to these gains as the frequency becomes very large and very small.

As omega to infty:

G_C to 0

G_R to 1.

As omega to 0:

G_C to 1

G_R to 0.

This shows that, if the output is taken across the capacitor, high frequencies are attenuated (rejected) and low frequencies are passed. Thus, the circuit behaves as a low-pass filter. If, though, the output is taken across the resistor, high frequencies are passed and low frequencies are rejected. In this configuration, the circuit behaves as a high-pass filter.

The range of frequencies that the filter passes is called its bandwidth. The point at which the filter attenuates the signal to half its unfiltered power is termed its cutoff frequency. This requires that the gain of the circuit be reduced to

G_C = G_R = frac{1}{sqrt{2}}.

Solving the above equation yields

omega_{c} = frac{1}{RC} mathrm{rad/s}

or

f_c = frac{1}{2pi RC} mathrm{Hz}

which is the frequency that the filter will attenuate to half its original power.

Clearly, the phases also depend on frequency, although this effect is less interesting generally than the gain variations.

As omega to 0:

phi_C to 0

phi_R to 90^{circ} = pi/2^{c}.

As omega to infty:

phi_C to -90^{circ} = -pi/2^{c}

phi_R to 0

So at DC (0 Hz), the capacitor voltage is in phase with the signal voltage while the resistor voltage leads it by 90°. As frequency increases, the capacitor voltage comes to have a 90° lag relative to the signal and the resistor voltage comes to be in-phase with the signal.

Time-domain considerations

The most straightforward way to derive the time domain behaviour is to use the Laplace transforms of the expressions for V_C and V_R given above. This effectively transforms jomega to s. Assuming a step input (i.e. V_{in} = 0 before t = 0 and then V_{in} = V afterwards):

These equations are for calculating the voltage across the capacitor and resistor respectively while the capacitor is charging; for discharging, the equations are vice-versa. These equations can be rewritten in terms of charge and current using the relationships C=Q/V and V=IR (see Ohm's law).

Thus, the voltage across the capacitor tends towards V as time passes, while the voltage across the resistor tends towards 0, as shown in the figures. This is in keeping with the intuitive point that the capacitor will be charging from the supply voltage as time passes, and will eventually be fully charged and form an open circuit.

These equations show that a series RC circuit has a time constant, usually denoted tau = RC being the time it takes the voltage across the component to either rise (across C) or fall (across R) to within 1/e of its final value. That is, tau is the time it takes V_C to reach V(1 - 1/e) and V_R to reach V(1/e).

The rate of change is a fractionalleft(1 - frac{1}{e}right) per tau. Thus, in going from t=Ntau to t = (N+1)tau, the voltage will have moved about 63.2 % of the way from its level at t=Ntau toward its final value. So C will be charged to about 63.2 % after tau, and essentially fully charged (99.3 %) after about 5tau. When the voltage source is replaced with a short-circuit, with C fully charged, the voltage across C drops exponentially with t from V towards 0. C will be discharged to about 36.8 % after tau, and essentially fully discharged (0.7 %) after about 5tau. Note that the current, I, in the circuit behaves as the voltage across R does, via Ohm's Law.

,!V_R = V_{in} - V_C
.
The first equation is solved by using an integrating factor and the second follows easily; the solutions are exactly the same as those obtained via Laplace transforms.

Integrator

Consider the output across the capacitor at high frequency i.e.

omega gg frac{1}{RC}.

This means that the capacitor has insufficient time to charge up and so its voltage is very small. Thus the input voltage approximately equals the voltage across the resistor. To see this, consider the expression for I given above:

I = frac{V_{in}}{R+1/jomega C}
but note that the frequency condition described means that

Parallel circuit

The parallel RC circuit is generally of less interest than the series circuit. This is largely because the output voltage V_{out} is equal to the input voltage V_{in} — as a result, this circuit does not act as a filter on the input signal unless fed by a current source.

With complex impedances:

I_R = frac{V_{in}}{R},
and

I_C = jomega C V_{in},
.

This shows that the capacitor current is 90° out of phase with the resistor (and source) current. Alternatively, the governing differential equations may be used:

I_R = frac{V_{in}}{R}
and

I_C = Cfrac{dV_{in}}{dt}
.

For a step input (which is effectively a 0 Hz or DC signal), the derivative of the input is an impulse at t=0. Thus, the capacitor reaches full charge very quickly and becomes an open circuit — the well-known DC behaviour of a capacitor.